# How do you use cos(t) and sin(t), with positive coefficients, to parametrize the intersection of the surfaces x^2+y^2=25 and z=3x^2?

Use the following parametrization for the curve $s$ generated by the intersection:

$s \left(t\right) = \left(x \left(t\right) , y \left(t\right) , z \left(t\right)\right) , t \in \left[0 , 2 \pi\right)$
$x = 5 \cos \left(t\right)$
$y = 5 \sin \left(t\right)$
$z = 75 {\cos}^{2} \left(t\right)$

Note that $s \left(t\right) : \mathbb{R} \to {\mathbb{R}}^{3}$ is a vector valued function of a real variable.

To reach this result, consider the curves that these equations define on certain planes.

The equation ${x}^{2} + {y}^{2} = 25$ defines a circle of radius $5$ centered on the $z$-axis on the planes $z = {c}_{1}$, where ${c}_{1} \in \mathbb{R}$ is any constant.
The equation $z = 3 {x}^{2}$ defines a parabola on any plane $y = {c}_{2}$, where ${c}_{2} \in \mathbb{R}$ is another constant.
The surfaces are, therefore, those obtained by translating the circle along the $z$-axis and the parabola along the $y$-axis.

To obtain a parametrization for the intersection curve $s$, we must find equations for $x$, $y$ and $z$ as functions of $t$ that obey both equations given in the problem.

Consider the standard parametrization for a circle $C$ of radius $r$ (it's easy to see that this parametrization fulfils the condition ${x}^{2} + {y}^{2} = {r}^{2}$):

$C \left(t\right) = \left(r \cos \left(t\right) , r \sin \left(t\right)\right) , t \in \left[0 , 2 \pi\right)$

Checking the first equation, we get $r = 5$ and

$x = 5 \cos \left(t\right)$
$y = 5 \sin \left(t\right)$

Now, we already have an expression for $x \left(t\right)$. So, in order to obey the second condition, we make:

$z = 3 {x}^{3} = 3 {\left(5 \cos \left(t\right)\right)}^{2} = 75 {\cos}^{2} \left(t\right)$

And we have the parametrization $s \left(t\right) = \left(x \left(t\right) , y \left(t\right) , z \left(t\right)\right) , t \in \left[0 , 2 \pi\right)$ for $s$.